mat25-hw1solns

mat25-hw1solns - Math 25: Advanced Calculus UC Davis,...

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Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 — Solutions to Homework Assignment #1 Remember throughout that there may be more than one correct way to prove any given result. Solutions 1. Let n be a natural number, divisible both by 5 and by 3. Because n is divisible by 5, there exists some natural number m such that n = 5 m . Since n is also divisible by 3, n 3 = 5 m 3 must be natural. Now, we use the twin facts that any natural multiple of a natural number is natural, and that any (nonnegative) difference between natural numbers must also be natural. Using the first, 2 m = 6 m 3 is natural; so, using the second, 6 m 3 - 5 m 3 = m 3 is natural. By substitution, then, ( n 5 ) 3 = n 15 is a natural number, and hence n is divisible by 15. 2. We disprove the claim directly, by offering a counterexample. Let n = 24. Then n 6 = 4 and n 8 = 3, so that n is divisible both by 6 and by 8. However,
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This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.

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mat25-hw1solns - Math 25: Advanced Calculus UC Davis,...

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